高数题目求助！！急急急！！！
展开全部
令x=tant
x=1,t=π/4
x=√3,t=π/3
dx=sec^2tdt
原式=∫[π/4,π/3] 1/(tan^2t*sect)*sec^2tdt
=∫[π/4,π/3] 1/(sintcost)dt
=∫[π/4,π/3] (sin^2t+cos^2t)/(sintcost)dt
=∫[π/4,π/3] (sint/cost+cost/sint)dt
=(-lncost+lnsint)[π/4,π/3]
=lntant[π/4,π/3]
=1/2ln3
算一下就可以了
x=1,t=π/4
x=√3,t=π/3
dx=sec^2tdt
原式=∫[π/4,π/3] 1/(tan^2t*sect)*sec^2tdt
=∫[π/4,π/3] 1/(sintcost)dt
=∫[π/4,π/3] (sin^2t+cos^2t)/(sintcost)dt
=∫[π/4,π/3] (sint/cost+cost/sint)dt
=(-lncost+lnsint)[π/4,π/3]
=lntant[π/4,π/3]
=1/2ln3
算一下就可以了
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
